Cho \(\dfrac{x_1}{x_2}=\dfrac{x_2}{x_3}=\dfrac{x_3}{x_4}=\dfrac{x_4}{x_5}=...=\dfrac{x_{2018}}{x_{2019}}.Chứng.minh:(\dfrac{x_1+x_2+x_3+...+x_{1018}}{x_2+x_3+x_4+...+x_{2019}})^{2018}=\dfrac{x_1}{x_{2019}}\)

Ai đó giúp mình nhé

cho \(\frac{_{x_1}}{x_2}=\frac{x_2}{x_3}=\frac{x_3}{x_4}=\frac{x_4}{x_5}=...=\frac{x_{2008}}{x_{2009}}\). Chứng minh rằng: \(\left(\frac{x_1+x_2+x_3+x_4+...+x_{2008}}{x_2+x_3+x_4+x_5+...+x_{2009}}\right)^{2008}\) = \(\frac{x_1}{x_{2009}}\)

Cho: 

\(\frac{x_1-1}{2017}=\frac{x_2-2}{2016}=\frac{x_3-3}{2015}=...=\frac{x_{2017}-2017}{1}vàx_1+x_2+...+x_{2017=2017\cdot2018.}Tìmx_1,x_2,x_{3,...,x_{2017}?}\)

cho \(x_1+x_2+x_3+...+x_{50}+x_{51}=0\)và\(x_1+x_2=x_3+x_4=x_5+x_6=...=x_{49}+x_{50}=1\)Tính \(x_{50}\)

Cho \(x_1+x_2+x_3+...+x_{50}+x_{51}=0\)

Và \(x_1+x_2=x_3+x_4=x_5+x_6=...=x_{49}+x_{50}=1\)

Tính \(x_{50}\)

Cảm ơn nhiều ạ!!!

Cho \(\left(2017x_1-2016y_1\right)^2+\left(2017x_2-2016y_2\right)^2+...+\left(2017x_{2016}-2016y_{2017}\right)^2\le0\)

CMR: \(\frac{x_1+x_2+x_3+...+x_{2016}}{u+y_1+y_2+y_3+...+y_{2016}}=\frac{2016}{2017}\)

tìm \(x_1,x_2,x_3.......,x_9\)

\(\frac{x_{1-1}}{9}=\frac{x_{2-2}}{8}=\frac{x_3-3}{7}=....=\frac{x_{9-9}}{1}\) và \(x_1+x_2+x_3+...+x_9=90\)

tìm \(x_1;x_2;x_3;......;x_{2011}\) biet 

\(\frac{x_1-1}{2010}=\frac{x_2-2}{2009}=.....=\frac{x_{2010}-2010}{1}\)va \(x_1+x_2+.....+x_{2011}=2\left(1+2+3+...+2010\right)\)